Mathematics, 05.03.2020 01:44 jholland03

Show that the area of this right triangle is one half bc sine Upper A 1 2bcsinA. In right triangle A B C, the side opposite vertex A is labeled a, the side opposite vertex B is labeled b, and the side opposite vertex C is labeled c. Angle C is a right angle. The area of this right triangle is ▼ one half ac. 1 2ac. one half ab. 1 2ab. one half bc. 1 2bc. The base a of the right triangle ABC is ▼ c sine A. csinA. 2 c sine A.2csinA. 2 a cosine B.2acosB. a cosine B. acosB. Substituting aequals=c sine Upper AcsinA into the area formula gives the area of this right triangle is ▼ bc sine A. bcsinA. one half bc cosine B. 1 2bc cos B. one half bc sine A. 1 2bcsinA. bc cosine B.

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Show that the area of this right triangle is one half bc sine Upper A 1 2bcsinA. In right triangle A...

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